Question

Cot A-1/2-sec^2 A= cot A/1+tan A

Answer 1

$\huge\tt\dag\underline\pink{Solution}:-$

$\tt\purple{\frac{Cot\:A-1}{2-sec²A}\:=\:\frac{cot\:A}{1+tanA}}~~~~~\tt\purple{taking\:L.S.H\:to\:prove}$

$\tt = \frac{ [\frac{1}{tan \: A} ]}{2 - (1 + tan {}^{2} A)} \\ \\ \tt( \therefore cot \: A \: = \frac{1}{tan \: A}) \\ \\ \tt (\therefore sec {}^{2} A = 1 + tan {}^{2} A) \\ \\ \\ = \tt \frac{ \frac{(1 - tan \: A)}{tan \: A} }{2 - 1 - tan {}^{2} A} \\ \\ \tt = \frac{ \frac{(1 - tan \: A)}{tan \: A} }{1 - tan {}^{2}a } \\ \\ \\ \tt( \therefore \: a {}^{2} - b {}^{2} = (a - b)(a + b)) \\ \\ \\ \tt = \: \frac{ \cancel{(1 - tan \: a)}}{tan \: A\: \cancel{ (1 - tan \: A)}(1 + tan \: A)} \\ \\ \\ \tt = \frac{1}{tan \: A\: (1 + tan \: A)} \\ \\ \\ = \tt \frac{cot \: A}{1 + tan \: A}$

$~~~~~~~~~~~~~~~~~~~~\large\sf \fbox\pink{Hence, proved}$